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I don't understand well this situation

Any associative magma (i.e., a semigroup) is alternative. More generally, a magma in which every pair of elements generates an associative submagma must be alternative. The converse, however, is not true, in contrast to the situation in alternative algebras. In fact, an alternative magma need not even be power-associative.

  1. Why an alternative magma need not even be power-associative ?
  2. When an alternative magma could be, instead, always any associative magma ?

power associativity is a property of a binary operation which is a weak form of associativity.

So in opposite direction should we have a more strong form of associativity to start from an alternative magma and return to any associative magma (that is for definition, alternative) ?

Jack
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1 Answers1

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I guess an answer to your first question can be to provide a counter-example.
Consider the magma with the following multiplication table:
enter image description here

It is easy to check that it is an alternative magma.
Indeed, in order to see that it satisfies $$(x \cdot x) \cdot y = x \cdot (x \cdot y)$$ and $$x \cdot (y \cdot y) = (x \cdot y) \cdot y,$$ we don't need to check with the element $1$, since it is neutral and yields trivial identities.
Also, this magma is commutative, and so the cases in which $x$ and $y$ are the same are also trivial.
So the only relevant cases to check are the pairs $(a,b)$, $(a,c)$ and $(b,c)$, all of these satisfying the required identities.

Now in this page you can see a proof that alternative magmas have well defined powers up to the fifth, that is, they satisfy $$x^2 \cdot x = x \cdot x^2,$$ $$x^3 \cdot x = x^2 \cdot x^2 = x \cdot x^3,$$ and $$x^4 \cdot x = x^3 \cdot x^2 = x^2 \cdot x^3 = x \cdot x^4,$$ and so we can use exponents up to $5$ in our computations.
Now, $a^2 = b$, whence $a^3 = ba = c$, yielding $$a^3 \cdot a^3 = c^2 = 1,$$ while, for example. $$a^2 \cdot a^4 = b \cdot b^2 = b \cdot a = c,$$ and so $a^6$ is not well defined, and therefore the magma is not power-associative.

I don't understand the content of your second question...

amrsa
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  • When you say that $a^6$ is not well defined, and therefore the magma is not power-associative can you say that is, instead, associative ? – Jack Mar 02 '19 at 17:49
  • @Jack No, because associative is stronger than power-associative. Check the definitions, if it were associative, then $$x^nx^m = (x_1\cdot\cdots\cdot x_n)\cdot(x_{n+1}\cdot\cdots\cdot x_{n+m}) = (x_1\cdot\cdots\cdot x_k)\cdot(x_{k+1}\cdot\cdots\cdot x_{n+m-k})=x^kx^{m+n-k},$$ where $x_i=x$, for $i=1,\ldots n+m$ and $0\leq k \leq m+n$. – amrsa Mar 02 '19 at 19:12
  • and also alternativity is stronger than power-associativity but weaker than associativity.. Exist labels, structure examples that shows a condition weaker still than power-associativity? – Jack Mar 03 '19 at 09:31
  • An algebra (or more generally a magma) is said to be power-associative if the subalgebra generated by any element is associative. But if the subalgebra generated by any element are alternative? this would be a weaker condition of the power-associativity .. but so .. we will go to infinity with definitions... – Jack Mar 03 '19 at 09:36
  • @Jack That's a different question. But I suspect that the subalgebra generated by an element being alternative is equivalent to the algebra being power-associative. I might be thinking wrong, but that's my first guess. Anyway, I'm glad my answer was helpful. Best wishes! – amrsa Mar 03 '19 at 10:18